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Perimeter of a Regular Polygon Inscribed in a Circle Radius r

Suppose a regular polygon with  
\[n\]
  sides is inscribed in a circle of radius  
\[r\]
, with the vertices of the polygon touching the circle.
From the centre of the circle lines are drawn to each vertex, and  
\[n\]
  triangles are formed. The angle of each triangle at the centre of the circle is  
\[\frac{360}{n} \equiv \frac{2 \pi}{n}\]
, and the length of the side opposite this angle is is
\[\begin{equation} \begin{aligned} a &= \sqrt{r^2+r^2-2 \times r \times r \times cos( \frac{2 \pi}{n}} \\ &= r \sqrt{2-2 cos( \frac{2 \pi}{n}} \\ &= r \sqrt{2} \sqrt{1-cos( \frac{2 \pi}{n}} \\ &= r \sqrt{2} \sqrt{2 sin^2 \frac{\pi}{n}} \\ &= 2 r sin \frac{\pi}{n} \end{aligned} \end{equation}\]
.

There are  
\[n\]
  such triangles, so the perimeter of the polygon is  
\[n \times 2 r sin \frac{\pi}{n} \]
.
As the number of sides gets larger  
\[sin (\frac{\pi}{n}) \]
  tends to  
\[\frac{\pi}{n}\]
, so the perimeter of the polygon tends to  
\[n \times 2r \frac{\pi}{n}=2 \pi r\]
.